Question

Question 12 of 19 (1 point) View problem in a pop-up 10.2 Section Exercise 13 Question Traiic accidents: Traffic engineers compared rates of traffic accidents at intersections with raised medians with rates at intersections with two-way left-turn lanes. They found that out of 4655 accidents at intersections with raised medians, 2281 were rear-end accidents, and out of 4585 accidents at two-way left-turn lanes, 2037 were rear-end accidents. Check Answ Solve It Guided Solutio Part 1 out of 2 Assuming these to be random samples of accidents from the two types of intersection, construct aShow Exampl 98% confidence interval for the difference between the proportions of accidents that are of the rear-end type at the two types of intersection. Let P1 denote the proportion of accidents of the rear-end type at intersections with raised medians. Use tables to find the critical value and round the answers to three decimal places. Link to Textboo Tables Formulas A 98% confidence interval for the difference between the proportions of accidents that are of the rear-end type at the two types of intersection is <pi-p2 < コ Save for Later Submit Assignment

Answer 1

12. The statistical software output for this problem is:

Two sample proportion summary confidence interval:
p₁ : proportion of successes for population 1
p₂ : proportion of successes for population 2
p₁ - p₂ : Difference in proportions

98% confidence interval results:

Difference	Count1	Total1	Count2	Total2	Sample Diff.	Std. Err.	L. Limit	U. Limit
p1 - p2	2281	4655	2037	4585	0.045735932	0.010369793	0.021612186	0.069859678

Hence,

98% confidence interval will be:

0.022 < p1 - p2 < 0.070